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Continuity at a point in terms of closure



The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Set Closure Union and IntersectionContinuity of a function through adherence of subsetsConvex set with empty interior is nowhere dense?Is preimage of closure equal to closure of preimage under continuous topological maps?How to show the logical equivalence of the following two definitions of continuity in a topological space?Given $A subseteq X$ in the discrete and the trivial topology, find closure of $A$Show two notions of dense are equivalentEquivalent definitions of continuity at a pointAbout continuity and clousureEquivalent definition of irreducible topological subspace.










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If $X$ and $Y$ are topological spaces, for $f:Xto Y$ to be continuous at $x_0in X$ it is necessary that $Asubseteq X land x_0inoverlineA implies f(x_0)inoverlinef(A)$.



I was wondering whether it is also sufficient. A proof or counterexample would be much appreciated!



Proof (necessity): Let $V$ be a neighborhood of $f(x_0)$. Since $f$ is continuous, $f^-1(V)$ is a neighborhood of $x_0$ in $X$. Since $x_0inoverlineA$, we have $Acap f^-1(V)neqvarnothing$. Let $xin Acap f^-1(V)$. Then $f(x)in f(A)cap V$, so that $f(A)cap Vneqvarnothing$; since this holds for any neighborhood $V$ of $f(x_0)$, we have $f(x_0)inoverlinef(A)$.










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    4












    $begingroup$


    If $X$ and $Y$ are topological spaces, for $f:Xto Y$ to be continuous at $x_0in X$ it is necessary that $Asubseteq X land x_0inoverlineA implies f(x_0)inoverlinef(A)$.



    I was wondering whether it is also sufficient. A proof or counterexample would be much appreciated!



    Proof (necessity): Let $V$ be a neighborhood of $f(x_0)$. Since $f$ is continuous, $f^-1(V)$ is a neighborhood of $x_0$ in $X$. Since $x_0inoverlineA$, we have $Acap f^-1(V)neqvarnothing$. Let $xin Acap f^-1(V)$. Then $f(x)in f(A)cap V$, so that $f(A)cap Vneqvarnothing$; since this holds for any neighborhood $V$ of $f(x_0)$, we have $f(x_0)inoverlinef(A)$.










    share|cite|improve this question









    $endgroup$














      4












      4








      4





      $begingroup$


      If $X$ and $Y$ are topological spaces, for $f:Xto Y$ to be continuous at $x_0in X$ it is necessary that $Asubseteq X land x_0inoverlineA implies f(x_0)inoverlinef(A)$.



      I was wondering whether it is also sufficient. A proof or counterexample would be much appreciated!



      Proof (necessity): Let $V$ be a neighborhood of $f(x_0)$. Since $f$ is continuous, $f^-1(V)$ is a neighborhood of $x_0$ in $X$. Since $x_0inoverlineA$, we have $Acap f^-1(V)neqvarnothing$. Let $xin Acap f^-1(V)$. Then $f(x)in f(A)cap V$, so that $f(A)cap Vneqvarnothing$; since this holds for any neighborhood $V$ of $f(x_0)$, we have $f(x_0)inoverlinef(A)$.










      share|cite|improve this question









      $endgroup$




      If $X$ and $Y$ are topological spaces, for $f:Xto Y$ to be continuous at $x_0in X$ it is necessary that $Asubseteq X land x_0inoverlineA implies f(x_0)inoverlinef(A)$.



      I was wondering whether it is also sufficient. A proof or counterexample would be much appreciated!



      Proof (necessity): Let $V$ be a neighborhood of $f(x_0)$. Since $f$ is continuous, $f^-1(V)$ is a neighborhood of $x_0$ in $X$. Since $x_0inoverlineA$, we have $Acap f^-1(V)neqvarnothing$. Let $xin Acap f^-1(V)$. Then $f(x)in f(A)cap V$, so that $f(A)cap Vneqvarnothing$; since this holds for any neighborhood $V$ of $f(x_0)$, we have $f(x_0)inoverlinef(A)$.







      general-topology continuity






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      asked Apr 7 at 16:29









      BlondCaféBlondCafé

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          Let $Vsubset Y$ be an open such that $f(x_0)in V$. If $x_0 in overlinef^-1(Ysetminus V)$, then $f(x_o)in overlinef(f^-1(Ysetminus V))subset overlineYsetminus V= Ysetminus V$, a contradiction, so $x_0 notin overlinef^-1(Ysetminus V)$. Then, if $U = Xsetminus overlinef^-1(Ysetminus V)$ is an open in $X$ such that $x_0in U$, and $f(U)subset V$, so $f$ is continuous in $x_0$.






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            $begingroup$

            It's also sufficient: let $y=f(x_0)$ and $y in V$, $V$ open in $Y$.
            We want to find (for continuity at $x_0$) find some open neighbourhood $U$ of $x_0$ such that $f[U] subseteq V$.



            Suppose that this would fail, then for every neighbourhood $U$ of $x_0$ we would have $f[U] nsubseteq V$, or equivalently $U cap f^-1[Ysetminus V] neq emptyset$.



            It follows that then $x_0 in overlinef^-1[Ysetminus V]$ and so the assumption on $f$ would imply that $y=f(x_0) in overlinef[f^-1[Ysetminus V]]$. But $f[f^-1[B]] subseteq B$ for any $B$ so we'd deduce that $y in overlineYsetminus V = Ysetminus V$ which is nonsense. So contradiction and such a $U$ must exist.






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              3












              $begingroup$

              Let $Vsubset Y$ be an open such that $f(x_0)in V$. If $x_0 in overlinef^-1(Ysetminus V)$, then $f(x_o)in overlinef(f^-1(Ysetminus V))subset overlineYsetminus V= Ysetminus V$, a contradiction, so $x_0 notin overlinef^-1(Ysetminus V)$. Then, if $U = Xsetminus overlinef^-1(Ysetminus V)$ is an open in $X$ such that $x_0in U$, and $f(U)subset V$, so $f$ is continuous in $x_0$.






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              $endgroup$

















                3












                $begingroup$

                Let $Vsubset Y$ be an open such that $f(x_0)in V$. If $x_0 in overlinef^-1(Ysetminus V)$, then $f(x_o)in overlinef(f^-1(Ysetminus V))subset overlineYsetminus V= Ysetminus V$, a contradiction, so $x_0 notin overlinef^-1(Ysetminus V)$. Then, if $U = Xsetminus overlinef^-1(Ysetminus V)$ is an open in $X$ such that $x_0in U$, and $f(U)subset V$, so $f$ is continuous in $x_0$.






                share|cite|improve this answer









                $endgroup$















                  3












                  3








                  3





                  $begingroup$

                  Let $Vsubset Y$ be an open such that $f(x_0)in V$. If $x_0 in overlinef^-1(Ysetminus V)$, then $f(x_o)in overlinef(f^-1(Ysetminus V))subset overlineYsetminus V= Ysetminus V$, a contradiction, so $x_0 notin overlinef^-1(Ysetminus V)$. Then, if $U = Xsetminus overlinef^-1(Ysetminus V)$ is an open in $X$ such that $x_0in U$, and $f(U)subset V$, so $f$ is continuous in $x_0$.






                  share|cite|improve this answer









                  $endgroup$



                  Let $Vsubset Y$ be an open such that $f(x_0)in V$. If $x_0 in overlinef^-1(Ysetminus V)$, then $f(x_o)in overlinef(f^-1(Ysetminus V))subset overlineYsetminus V= Ysetminus V$, a contradiction, so $x_0 notin overlinef^-1(Ysetminus V)$. Then, if $U = Xsetminus overlinef^-1(Ysetminus V)$ is an open in $X$ such that $x_0in U$, and $f(U)subset V$, so $f$ is continuous in $x_0$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Apr 7 at 17:06









                  guchiheguchihe

                  21919




                  21919





















                      2












                      $begingroup$

                      It's also sufficient: let $y=f(x_0)$ and $y in V$, $V$ open in $Y$.
                      We want to find (for continuity at $x_0$) find some open neighbourhood $U$ of $x_0$ such that $f[U] subseteq V$.



                      Suppose that this would fail, then for every neighbourhood $U$ of $x_0$ we would have $f[U] nsubseteq V$, or equivalently $U cap f^-1[Ysetminus V] neq emptyset$.



                      It follows that then $x_0 in overlinef^-1[Ysetminus V]$ and so the assumption on $f$ would imply that $y=f(x_0) in overlinef[f^-1[Ysetminus V]]$. But $f[f^-1[B]] subseteq B$ for any $B$ so we'd deduce that $y in overlineYsetminus V = Ysetminus V$ which is nonsense. So contradiction and such a $U$ must exist.






                      share|cite|improve this answer









                      $endgroup$

















                        2












                        $begingroup$

                        It's also sufficient: let $y=f(x_0)$ and $y in V$, $V$ open in $Y$.
                        We want to find (for continuity at $x_0$) find some open neighbourhood $U$ of $x_0$ such that $f[U] subseteq V$.



                        Suppose that this would fail, then for every neighbourhood $U$ of $x_0$ we would have $f[U] nsubseteq V$, or equivalently $U cap f^-1[Ysetminus V] neq emptyset$.



                        It follows that then $x_0 in overlinef^-1[Ysetminus V]$ and so the assumption on $f$ would imply that $y=f(x_0) in overlinef[f^-1[Ysetminus V]]$. But $f[f^-1[B]] subseteq B$ for any $B$ so we'd deduce that $y in overlineYsetminus V = Ysetminus V$ which is nonsense. So contradiction and such a $U$ must exist.






                        share|cite|improve this answer









                        $endgroup$















                          2












                          2








                          2





                          $begingroup$

                          It's also sufficient: let $y=f(x_0)$ and $y in V$, $V$ open in $Y$.
                          We want to find (for continuity at $x_0$) find some open neighbourhood $U$ of $x_0$ such that $f[U] subseteq V$.



                          Suppose that this would fail, then for every neighbourhood $U$ of $x_0$ we would have $f[U] nsubseteq V$, or equivalently $U cap f^-1[Ysetminus V] neq emptyset$.



                          It follows that then $x_0 in overlinef^-1[Ysetminus V]$ and so the assumption on $f$ would imply that $y=f(x_0) in overlinef[f^-1[Ysetminus V]]$. But $f[f^-1[B]] subseteq B$ for any $B$ so we'd deduce that $y in overlineYsetminus V = Ysetminus V$ which is nonsense. So contradiction and such a $U$ must exist.






                          share|cite|improve this answer









                          $endgroup$



                          It's also sufficient: let $y=f(x_0)$ and $y in V$, $V$ open in $Y$.
                          We want to find (for continuity at $x_0$) find some open neighbourhood $U$ of $x_0$ such that $f[U] subseteq V$.



                          Suppose that this would fail, then for every neighbourhood $U$ of $x_0$ we would have $f[U] nsubseteq V$, or equivalently $U cap f^-1[Ysetminus V] neq emptyset$.



                          It follows that then $x_0 in overlinef^-1[Ysetminus V]$ and so the assumption on $f$ would imply that $y=f(x_0) in overlinef[f^-1[Ysetminus V]]$. But $f[f^-1[B]] subseteq B$ for any $B$ so we'd deduce that $y in overlineYsetminus V = Ysetminus V$ which is nonsense. So contradiction and such a $U$ must exist.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Apr 7 at 17:07









                          Henno BrandsmaHenno Brandsma

                          116k349127




                          116k349127



























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